1.4 Rational Numbers 29(MO) Let x,y E Qp. Then 3m,n,m',n' E Zp 3 n , n' I= 0, x =~ and
y -- m' ti!• Th us,
mm' mm'
x · y = - · - = --E Qp. (Justify)
n n' nn'
(Ml) This property is "inherited" from F. (Explain)
(M2) Inherited from F. (Explain)
(M3) lE Qp, since 1 = t· This element of «J!F satisfies the condition
specified in (M3).
(M4) Let x E Qp 3 x I= 0. Then 3m,n E Zp 3 m,n,I= 0, and x = ~
Then x -^1 = ~ E Qp (Justify). This element x-^1 satisfies the property required
by (M4).
(D) Inherit ed from F. (Explain)
(01)- (03) are also inherited from F. (Explain) •Definition 1.4.4 If an ordered field F contains an element that is not a ra-
tional number (by our definition) then such an element is called an irrational
element of F.
Theorem 1.4.5 There is no element of «J!F whose square is 2.
Proof. For contradiction, suppose 3 x E Qp 3 x^2 = 2. Then 3 m, n E Zp 3
n I= 0 and x = ~. Without loss of generality, we may assume that m and n
have no prime factors in common. Then we have
(:)2 = 2,so
m^2 = 2n^2.Thus, m^2 is divisible by 2. By Exercise 3 below, this means that m is
divisible by 2. That is, 3 k E Zp 3 m = 2k. Thus, the above equation becomes(2k)^2 =2n^2 , so
4k^2 = 2n^2 , so 2k^2 = n^2.But this means that n^2 is even, which means that n is also even. Thus the
numerator, m, and the denominator, n, are both divisible by 2. Contradiction.
Hence, ~x E Qp 3 x^2 = 2. •SIMPLIFYING THE NOTATION
To simplify notation, we shall discontinue using the notation Zp and Qp,
and from now on simply use Z and Q! to denote these sets. We can do this
without ambiguity, because the symbols 1, 2, 3, ... mean the same in all ordered
fields.